Supermartingales and convergence

Question

These feel like basic enough questions, but I don't know where to find the answer.

Let $X_1,X_2,X_3,\dots$ be a supermartingale such that $|X_{n+1} - X_n| < K$ for all $n$ ($K$ fixed). Does the event $X_n \rightarrow +\infty$ necessarily have probability zero? What if we also have the condition that given $X_1,\dots,X_n$ there are only $N$ possibilities for $X_{n+1}$, each with probability $1/N$?

Suppose in addition that $\mathbb{E}(X_{n+1}|X_1,\dots,X_n) \le X_n - c_n$, where $\sum c_n \rightarrow +\infty$. Is $X_n \rightarrow -\infty$ almost certain?

Answer 1

For a martingale $M_n$ with bounded increment, then, almost surely :

• either $M_n$ converges to a finite limite.

• or $\limsup M_n=\infty$ and $\liminf M_n=-\infty$.

Sketch of the proof. One can assume that $M_0=0$. Let $T$ be the first time at which the martingale goes below $-A$. Then $M_{n \wedge N}$ is a martingale bounded from below, therefore it converges. Then $M_n$ converges if $N$ is infinite etc.

For your problem consider the Doob decomposition.

Answer 2

For the second part of the question: it depends on the mean and variance you have.

Let $X_t=M_t+Y_t$ where $M_t$ is a martingale and $Y_t$ is predictable. Let $\mu_t=\mathbb E[X_t]=\mathbb E[Y_t]$ and $\sigma_t^2=\sum_{s\le t}\mathbb E[(M_s-M_{s-1})^2|M_1,\ldots,M_{s-1}]$. By martingale central limit theorem, $M_t$ is asymptotically normal with mean 0 and variance $\sigma_t^2$, so a sufficient condition for $X_t\rightarrow-\infty$ almost surely is $\sigma_t=o(|\mu_t|)$ as $t\rightarrow\infty$.

(To be precise, $\sigma_t$ is a random variable, so you really need to do some fiddling with stopping times as demonstrated in the Wikipedia article on martingale CLT, but the gist of it is that it will converge almost surely if the mean dominates the variance.)